# If there are (2n +1) terms in A.P., then prove that the ratio of the sum of odd terms and the sum of even terms is (n + 1):n.

Answer

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Hint: First, analyse the odd and even terms separately from an A.P with common difference, from that of the original A.P. Compute the sum of odd and even terms and find the ratio in terms of n.

Complete step-by-step answer:

The A.P. containing 2n + 1 terms can be divided into two different A.P.s by separating the odd and even terms of the original A.P. The A.P. formed by taking the odd terms will have (n + 1) terms and the A.P. formed by taking the even terms will have n terms.

Let the first term of the original A.P. be a and the common difference be d. Then, the first odd term is a and the first even term is a + d.

We know the formula for the \[{n^{th}}\] term of the A.P.

\[{t_n} = a + (n - 1)d........(1)\]

The last even term of the A.P. is the \[2{n^{th}}\] term and it is given by:

\[{l_{even}} = a + (2n - 1)d.........(2)\]

The last odd term of the A.P. is the \[{(2n + 1)^{th}}\] term and it is given by:

\[{l_{odd}} = a + (2n + 1 - 1)d\]

\[{l_{odd}} = a + 2nd...........(3)\]

We know that the sum to n terms of the A.P. is given as:

\[{S_n} = \dfrac{n}{2}(a + l)............(4)\]

The sum of odd terms of the A.P. is given as:

\[{S_{odd}} = {S_{n + 1}} = \dfrac{{n + 1}}{2}(a + {l_{odd}})\]

From equation (3), we get:

\[{S_{odd}} = \dfrac{{n + 1}}{2}(a + a + 2nd)\]

\[{S_{odd}} = \dfrac{{n + 1}}{2}(2a + 2nd)\]

Taking 2 out as common and cancelling with the denominator, we have:

\[{S_{odd}} = (n + 1)(a + nd).........(5)\]

The sum of even terms of the A.P. is given as:

\[{S_{even}} = {S_n} = \dfrac{n}{2}(a + d + {l_{even}})\]

Substituting equation (2) in the above equation, we have:

\[{S_{even}} = \dfrac{n}{2}(a + d + a + (2n - 1)d)\]

\[{S_{even}} = \dfrac{n}{2}(2a + 2nd)\]

Taking 2 out as common and cancelling with the denominator, we have:

\[{S_{even}} = n(a + nd).........(6)\]

Dividing equation (5) by equation (6), we have:

\[\dfrac{{{S_{odd}}}}{{{S_{even}}}} = \dfrac{{(n + 1)(a + nd)}}{{n(a + nd)}}\]

Cancelling (a + nd), we have:

\[\dfrac{{{S_{odd}}}}{{{S_{even}}}} = \dfrac{{n + 1}}{n}\]

Hence, we proved that the ratio of the sum of odd terms and the sum of even terms is (n + 1):n

Note: You must be careful when you choose the first even terms of the A.P, it is a + d and not a. The number of odd terms is (n + 1) and not n.

Complete step-by-step answer:

The A.P. containing 2n + 1 terms can be divided into two different A.P.s by separating the odd and even terms of the original A.P. The A.P. formed by taking the odd terms will have (n + 1) terms and the A.P. formed by taking the even terms will have n terms.

Let the first term of the original A.P. be a and the common difference be d. Then, the first odd term is a and the first even term is a + d.

We know the formula for the \[{n^{th}}\] term of the A.P.

\[{t_n} = a + (n - 1)d........(1)\]

The last even term of the A.P. is the \[2{n^{th}}\] term and it is given by:

\[{l_{even}} = a + (2n - 1)d.........(2)\]

The last odd term of the A.P. is the \[{(2n + 1)^{th}}\] term and it is given by:

\[{l_{odd}} = a + (2n + 1 - 1)d\]

\[{l_{odd}} = a + 2nd...........(3)\]

We know that the sum to n terms of the A.P. is given as:

\[{S_n} = \dfrac{n}{2}(a + l)............(4)\]

The sum of odd terms of the A.P. is given as:

\[{S_{odd}} = {S_{n + 1}} = \dfrac{{n + 1}}{2}(a + {l_{odd}})\]

From equation (3), we get:

\[{S_{odd}} = \dfrac{{n + 1}}{2}(a + a + 2nd)\]

\[{S_{odd}} = \dfrac{{n + 1}}{2}(2a + 2nd)\]

Taking 2 out as common and cancelling with the denominator, we have:

\[{S_{odd}} = (n + 1)(a + nd).........(5)\]

The sum of even terms of the A.P. is given as:

\[{S_{even}} = {S_n} = \dfrac{n}{2}(a + d + {l_{even}})\]

Substituting equation (2) in the above equation, we have:

\[{S_{even}} = \dfrac{n}{2}(a + d + a + (2n - 1)d)\]

\[{S_{even}} = \dfrac{n}{2}(2a + 2nd)\]

Taking 2 out as common and cancelling with the denominator, we have:

\[{S_{even}} = n(a + nd).........(6)\]

Dividing equation (5) by equation (6), we have:

\[\dfrac{{{S_{odd}}}}{{{S_{even}}}} = \dfrac{{(n + 1)(a + nd)}}{{n(a + nd)}}\]

Cancelling (a + nd), we have:

\[\dfrac{{{S_{odd}}}}{{{S_{even}}}} = \dfrac{{n + 1}}{n}\]

Hence, we proved that the ratio of the sum of odd terms and the sum of even terms is (n + 1):n

Note: You must be careful when you choose the first even terms of the A.P, it is a + d and not a. The number of odd terms is (n + 1) and not n.

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